l

GOALS1. Define a what is meant by a point estimate.2. Define the term level of confidence.3. Construct a confidence interval for the population

mean when the population standard deviation is known.

4. Construct a confidence interval for the population mean when the population standard deviation is unknown.

5. Construct a confidence interval for the population proportion.

6. Determine the sample size for attribute and variable sampling.

Chapter NineEstimation and Confidence IntervalsEstimation and Confidence Intervals

Point and Interval Estimates

A point estimate is a single value (statist ic) used to estimate a population value (parameter).

A interval estimate is a range of values within which the population parameter is expected to occur. The interval within which a population

parameter is expected to occur is called a confidence interval.

The specif ied probabil i ty is called the level of confidence.

The two confidence intervals that are used extensively are the 95% and the 99%.

Is the Population Normal

Is n 30 or more ? Is the population standard deviation known ?

Use a Non Parametric Test

Use the Z Distribution

Use the t Distribution

Use the Z Distribution

No

NoNo

Yes

YesYes

Interval Estimation for The Population Mean

Confidence Intervals

The degree to which we can rely on the statistic is as important as the init ial calculation. Remember, most of the time we are working from samples. And samples are really estimates. Ultimately, we are concerned with the accuracy of the estimate.

1. Confidence interval provides Range of Values Based on Observations from 1 Sample

1. Confidence interval gives Information about Closeness to Unknown Population Parameter Stated in terms of Probabil i ty Knowing Exact Closeness Requires Knowing

Unknown Population Parameter

Areas Under the Normal Curve

Between:± 1 σ - 68.26%± 2 σ - 95.44%± 3 σ - 99.74%

µµ-1σ µ+1σ

µ-2σ µ+2σµ+3σµ-3σ

I f we draw an observation from the normal distributed population, the drawn value is l ikely (a chance of 68.26%) to l ie inside the interval of (µ-1σ, µ+1σ).

P((µ-1σ <x<µ+1σ) =0.6826.

Elements of Confidence Interval Estimation

Confidence Interval

Sample Statistic

Confidence Limit (Lower)

Confidence Limit (Upper)

A probabil i ty that the population parameter fal ls somewhere within the interval.

Confidence Intervals

90% Samples

95% Samples

99% Samples

σx_

nZX

XZX σσ ⋅±=⋅±

X

XXXXσµσµµσµσµ ⋅+⋅+⋅−⋅− 58.2645.1645.158.2

XXσµσµ ⋅+⋅− 96.196.1

Level of Confidence

1. Probabil i ty that the unknown population parameter falls within the interval

2. Denoted (1 - α) % = level of confidence α Is the Probabil i ty That the Parameter Is

Not Within the Interval1. Typical Values Are 99%, 95%, 90%

Interpreting Confidence Intervals

Once a confidence interval has been constructed, it wil l either contain the population mean or it wil l not.

For a 95% confidence interval, i f you were to produce all the possible confidence intervals using each possible sample mean from the population, 95% of these intervals would contain the population mean.

Intervals & Level of Confidence

Sampling Distr ibution of Mean

Large Number of Intervals

Intervals Extend from

(1 - α) % of Intervals Contain µ .

α % Do Not.

µx = µ

1 - α α/2α/2

X_

σx_

XZX

XZX

σ

σ

⋅+

⋅− to

Point Estimates and Interval Estimates

The factors that determine the width of a confidence interval are:1. The size of the sample (n) from which the

statist ic is calculated.2. The variabil i ty in the population, usually

estimated by s.3. The desired level of confidence.

nZX

XZX σ

ασα ⋅±=⋅± )2/

()2/

(

Point and Interval Estimates

I f the population standard deviation is known or the sample is greater than 30 we use the z distr ibution.

n

szX ±

CONTOH Penelit ian dilakukan untuk mengetahui

pendapatan bersih PKL di Surabaya. Dari 100 orang sampel random diketahui rata-rata pendapatan bersih per hari PKL Rp 50.000 dengan simpangan baku RP 15.000. Berdasarkan data tersebut lakukan estimasi pendapatan bersih PKL di Surabaya dengan t ingkat keyakinan 95%.

Point and Interval Estimates

I f the population standard deviation is unknown and the sample is less than 30 we use the t distr ibution.

n

stX ±

Student’s t-Distribution

The t-distr ibution is a family of distr ibutions that is bell-shaped and symmetric l ike the standard normal distribution but with greater area in the tails. Each distr ibution in the t-family is defined by its degrees of freedom. As the degrees of freedom increase, the t-distr ibution approaches the normal distr ibution.

About Student

Student is a pen name for a stat ist ician named Will iam S. Gosset who was not allowed to publish under his real name. Gosset assumed the pseudonym Student for this purpose. Student’s t distr ibution is not meant to reference anything regarding college students.

Zt

0

t (df = 5)

Standard Normal

t (df = 13)Bell-Shaped

Symmetric

‘Fatter’ Tails

Student’s t-Distribution

Upper Tail Area

df .25 .10 .05

1 1.000 3.078 6.314

2 0.817 1.886 2.920

3 0.765 1.638 2.353

t0

Student’s t Table

Assume:n = 3df = n - 1 = 2α = .10α/2 =.05

2.920t Values

α / 2

.05

Degrees of freedom

Degrees of freedom refers to the number of independent data values available to estimate the population’s standard deviat ion. I f k parameters must be estimated before the population’s standard deviat ion can be calculated from a sample of size n, the degrees of freedom are equal to n - k.

Degrees of Freedom (df )

1. Number of Observations that Are Free to Vary After Sample Statistic Has Been Calculated

2. ExampleSum of 3 Numbers Is 6

X 1 = 1 (or Any Number)X 2 = 2 (or Any Number)X 3 = 3 (Cannot Vary)Sum = 6

degrees of freedom = n -1 = 3 -1= 2

t-Values

where:= Sample mean= Population mean

s = Sample standard deviationn = Sample size

n

sx

tµ−=

xµ

Estimation Example Mean (σ Unknown)

A random sample of n = 25 has = 50 and S = 8. Set up a 95% confidence interval estimate for µ.

X tS

nX t

S

nn n− ⋅ ≤ ≤ + ⋅

− ⋅ ≤ ≤ + ⋅

≤ ≤

− −α αµ

µ

µ

/ , / ,

. .

. .

2 1 2 1

50 2 06398

2550 2 0639

8

2546 69 53 30

X

Central Limit Theorem

For a population with a mean µ and a variance σ 2 the sampling distr ibution of the means of al l possible samples of size n generated from the population wil l be approximately normally distr ibuted.

The mean of the sampling distr ibution equal to µ and the variance equal to σ 2/n.

),?(~ 2σµX

)/,(~ 2 nNXn σµThe sample mean of n observation

The population distr ibution

Standard Error of the Sample Means

The standard error of the sample mean is the standard deviation of the sampling distr ibution of the sample means.

I t is computed by

is the symbol for the standard error of the sample mean.

σ is the standard deviation of the population.

n is the size of the sample.

nx

σσ =

xσ

Standard Error of the Sample Means

I f σ is not known and n ≥ 30, the standard deviation of the sample, designated s, is used to approximate the population standard deviation. The formula for the standard error is:

n

ssx =

95% and 99% Confidence Intervals for the sample mean

The 95% and 99% confidence intervals are constructed as follows:95% CI for the sample mean is given by

n

s96.1±µ

n

s58.2±µ

99% CI for the sample mean is given by

95% and 99% Confidence Intervals for µ

The 95% and 99% confidence intervals are constructed as follows:95% CI for the population mean is given by

n

sX 96.1±

n

sX 58.2±

99% CI for the population mean is given by

Constructing General Confidence Intervals for µ

In general, a confidence interval for the mean is computed by:

n

szX ±

EXAMPLE 3

The Dean of the Business School wants to estimate the mean number of hours worked per week by students. A sample of 49 students showed a mean of 24 hours with a standard deviation of 4 hours. What is the population mean?

The value of the population mean is not known. Our best estimate of this value is the sample mean of 24.0 hours. This value is called a point estimate.

Example 3 continued

Find the 95 percent confidence interval for the population mean.

12.100.2449

496.100.2496.1

±=

±=±n

sX

The confidence l imits range from 22.88 to 25.12.About 95 percent of the similarly constructed intervals include the population parameter.

Confidence Interval for a Population Proportion

The confidence interval for a population proportion is estimated by:

n

ppzp

)1( −±

EXAMPLE 4

A sample of 500 executives who own their own home revealed 175 planned to sell their homes and ret ire to Arizona. Develop a 98% confidence interval for the proport ion of executives that plan to sell and move to Arizona.

0497.35. 500

)65)(.35(.33.235. ±=±

CONTOH Lembaga riset melakukan penelit ian tentang

perusahaan di Jawa Timur yang sudah menerapkan UMR. Data menunjukkan dari 50 sampel perusahaan, 40 diantaranya sudah memenuhi UMR. Buatlah confidence interval 90% untuk menduga persentase perusahaan yang sudah menerapkan UMR.

Finite-Population Correction Factor

A population that has a f ixed upper bound is said to be f inite.

For a f inite population, where the total number of objects is N and the size of the sample is n, the following adjustment is made to the standard errors of the sample means and the proport ion:

Standard error of the sample means:

1−−=

N

nN

nx

σσ

Finite-Population Correction Factor

Standard error of the sample proportions:

1

)1(

−−−=

N

nN

n

pppσ

This adjustment is called the f inite-population correction factor.

I f n/N < .05, the f inite-population correction factor is ignored.

Finite-Population Correction Factor

Standard error of the sample proport ions:

This adjustment is called the f inite-population correction factor.

Note : I f n/N < 0.05, the f inite-population correction factor is ignored.

Interval Estimation for proport ion with f inite-pop

σp

p p

n

N n

N=

− −−

( )1

1

1

)ˆ1(ˆˆ

−−−±=

N

nN

n

ppZpP

EXAMPLE 5

Given the information in EXAMPLE 4, construct a 95% confidence interval for the mean number of hours worked per week by the students if there are only 500 students on campus.

Because n/N = 49/500 = .098 which is greater than 05, we use the f inite population correction factor.

0648.100.24)1500

49500)(

49

4(96.124 ±=

−−±

CONTOH Pimpinan bank ingin mengetahui tentang

kepuasan nasabah terhadap pelayanan bank. Dari jumlah nasabah 1000 orang, diambil sampel 100 orang untuk diwawancarai, Hasilnya 60 orang mengakui puas dengan pelayanan bank tersebut. Dengan α = 5%, berapa proporsi nasabah yang puas dengan pelayanan bank,

Selecting a Sample Size

There are 3 factors that determine the size of a sample, none of which has any direct relat ionship to the size of the population. They are:The degree of confidence selected. The maximum allowable error.The variat ion in the population.

Selecting a Sample Size

To find the sample size for a variable:

where : E is the allowable error, z is the z- value corresponding to the selected level of confidence, and s is the sample deviat ion of the pilot survey.

2*

*

=⇒=E

sznE

n

sz

nZX

XZX σ

ασα ⋅±=⋅± )2/

()2/

(

EXAMPLE 6

A consumer group would l ike to estimate the mean monthly electricity charge for a single family house in July within $5 using a 99 percent level of confidence. Based on similar studies the standard deviation is estimated to be $20.00. How large a sample is required?

1075

)20)(58.2(2

=

=n

Sample Size for Proportions

The formula for determining the sample size in the case of a proportion is:

where p is the estimated proport ion, based on past experience or a pilot survey; z is the z value associated with the degree of confidence selected; E is the maximum allowable error the researcher wil l tolerate.

2

)1(

−=E

Zppn

EXAMPLE 7

The American Kennel Club wanted to estimate the proportion of children that have a dog as a pet. I f the club wanted the estimate to be within 3% of the population proportion, how many children would they need to contact? Assume a 95% level of confidence and that the club estimated that 30% of the children have a dog as a pet.

89703.

96.1)70)(.30(.

2

=

=n

Two-sample Estimation Mean :

n 1, n 2 ≥ 30

n 1, n 2 < 30

Proport ion :

( )

+±−=−

2

22

1

21

2121 n

s

n

sZXXµµ

( )

+

−+

−+−±−=−2121

222

211

2121

11

2

)1()1(

nnnn

snsntXXµµ

( )2

22

1

112121

)ˆ1(ˆ)ˆ1(ˆˆˆ

n

pp

n

ppZppPP

−+−±−=−

Contoh Perusahaan ban sedang memebandingkan daya

pakai antara ban merek A dan Merek B. Dari sampel random 10 ban A diketahui rata-rata daya pakai 1.000 km dengan standar deviasi 100 km sedangkan dari sampel random 10 ban merek B rata-rata daya pakai 900 km dengan standar deviasi 90 km. Hitung perbedaan daya pakai antara ban merek A dan merek B dengan α = 0,05.

Contoh

Sampel random menunjukkan dari 80 kendaraan di kota A, 60 diantaranya telah melunasi pajak sedangkan di kota B dari 70 kendaraan, 40 diantaranya telah melunasi pajak. Hitunglah perbedaan persentase pelunasan pajak kendaraan di kedua kota tersebut dengan tingkat keyakinan 95%.

- END -

Chapter NineEstimation and Confidence IntervalsEstimation and Confidence Intervals